How many ways are there to make a sundae with two toppings if double toppings are allowed? An ice cream parlor has 6 toppings to choose from (hot fudge, sprinkles, caramel, cherries, blueberry sauce, and oreos. 
How many ways are there to make a sundae with two different toppings?
So for this question I believe its a combination and the number of possibilities would be 6C2. 
How many ways are there to make a sundae with two toppings if double toppings are allowed?
So since each topping has the option of being doubled would it be 12C2? I'm not really sure how to go about this. 
What would happen if there were three toppings with triple toppings allowed?
 A: In the two-toppings with repetition case, the answer is $\binom{7}{2}$. You want the number of mappings from 2 balls to a 5 boxes, up to permutations of the balls. The general formula for $n$ balls and $x$ boxes is $\binom{x+n-1}{n}$ 
This can be counted with stars and bars: You have 7 positions $*******$, and are choosing 2 of them to be the "stars"/balls, and the remaining 5 delimit the boxes. One such placement is $||*||*|$. If a star appears before the first bar, the first topping is used. If a star appears between the first and second bars, the second topping is used.... If a star appears after all of the bars, the 6th topping is used.
If 2 stars appear in any of the gaps between bars, that topping is used twice. 
Three toppings with triple toppings allowed can be counted in the same way: $n=3$ and $x=6$, so your number of choices is $\binom{6+3-1}{3}=\binom{8}{3}$.
A: With two different toppings and six to choose from you have a choice of six to go on first followed by five for the second, making a total of $6\times5=30$.
If you are allowed to repeat the same topping, the total is $6\times6=36$.
With three different toppings, the total is $6\times5\times4$ and if repeats are allowed, it's $6\times6\times6$
A: How many decisions are being made?  In this situation, 2 (first scoop and second scoop).
How many possibilities for the first scoop? - 6
How many possibilities for the second scoop? - 6
6x6 = 36 possible combinations, when this is treated as a permutation where the order matters.  For example, this solution suggests Sprinkles bottom - Cherry top is different from Cherry top - Sprinkles bottom.
If the order does not matter, then you would divide by 2, because each pattern has a reverse order that is a duplicate.
For three scoops and three toppings:
3 Decisions, where each decision has 3 possibilities.  This explanation again treats Cherry-Cherry-Sprinkles as different from Cherry-Sprinkles-Cherry.
First Choice: 3 Possible
Second Choice: 3 Possible
Third Choice: 3 Possible
3x3x3 = 27 Possible Permutations with Replacement
Permutation with Replacement --> P^R(n,r)=n^r where n is the number of choices (toppings) and r is the number of decisions (scoops).
Combination with Replacement --> (n+r−1)!/r!(n−1)! where n is the number of choices (toppings) and r is the number of decisions (scoops).
